Eh post a pic of the problem, I cant see it like that.
Yeah, but how did i get that?but its correct
Does u.x mean u times x? That's what I'm assuming.You must be registered for see images
Yeah, but how did i get that?
Are you doing derivatives through U substitution and partial derivatives?
Does u.x mean u times x? That's what I'm assuming.
-It looks like you're doing U substitution for y/x and then doing an integral(reverse derivative) of y' to find y=u*x
The question is "solve: y' = 1 + y/x"In which variable's respect are you differentiating the said equation?
Alright if U=y/x that means U'=1/x (I did the derivative of U with respect to x) then this puts y' as y'=u'*x+u put back in what you subbed out you'll have y'=(1/x)*x+(y/x). Then (1/x)*x equals 1 so in the end your back to y'=1+(y/x)U is subsititution for y/x, so u=y/x from that you can easily see that y=u*x
but what I don't get is how did I get the last part, how did I get that y'=u'*x+u
I still don't get it if you derivate y/x won't the outcome be -y*x^(-2)? how come it's 1/x?Alright if U=y/x that means U'=1/x (I did the derivative of U with respect to x) then this puts y' as y'=u'*x+u put back in what you subbed out you'll have y'=(1/x)*x+(y/x). Then (1/x)*x equals 1 so in the end your back to y'=1+(y/x)
-So to sum in up you subbed y/x to equal U
-Then you did the derivative of U to get 1/x equals U'
-put both U and U' back into the equation and you're set
-It depends on what you're holding constant or in other words what variable are you differentiating.I still don't get it if you derivate y/x won't the outcome be -y*x^(-2)? how come it's 1/x?
Do you need help with this too?The question is "solve: y' = 1 + y/x"
the answer is y = x ln |x| + C x
but I don't get that one step, no idea how to get y'=u'*x+u
Nah, the rest is fine, thanks.Do you need help with this too?